What is the last 2 digits of 2^(613)


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Question Number 14564 by tawa tawa last updated on 02/Jun/17

$What is the last 2 digits of 2^{613}$
Commented by tawa tawa last updated on 02/Jun/17

$God bless you sir .$
Commented by tawa tawa last updated on 02/Jun/17

$But the question is last two digit$
Commented by Tinkutara last updated on 02/Jun/17

$2 \equiv 2 (\mod 100)$ $2^{2} \equiv 4 (\mod 100)$ $2^{4} \equiv 16 (\mod 100)$ $2^{8} \equiv - 44 (\mod 100)$ $2^{16} \equiv 36 (\mod 100)$ $2^{32} \equiv - 4 (\mod 100)$ $2^{64} \equiv 16 (\mod 100)$ $2^{128} \equiv - 44 (\mod 100)$ $2^{256} \equiv 36 (\mod 100)$ $2^{512} \equiv - 4 (\mod 100)$ $2^{613} = 2^{512 + 64 + 32 + 4 + 1}$ $= (- 4) (16) (- 4) (16) (2) = 8192$ $\equiv 92 (\mod 100)$ $Hence last two digits are 92 .$
Commented by Tinkutara last updated on 02/Jun/17

$2 \equiv 2 (\mod 1000)$ $2^{2} \equiv 4 (\mod 1000)$ $2^{4} \equiv 16 (\mod 1000)$ $2^{8} \equiv 256 (\mod 1000)$ $2^{16} \equiv - 464 (\mod 1000)$ $2^{32} \equiv 296 (\mod 1000)$ $2^{64} \equiv - 384 (\mod 1000)$ $2^{128} \equiv 456 (\mod 1000)$ $2^{256} \equiv - 64 (\mod 1000)$ $2^{512} \equiv 96 (\mod 1000)$ $2^{613} = 2^{512 + 64 + 32 + 4 + 1}$ $= (96) (- 384) (296) (16) (2)$ $\equiv - 808 (\mod 1000)$ $\equiv 192 (\mod 1000)$ $Last 3 digits are 192 .$
Commented by tawa tawa last updated on 02/Jun/17

$i really appreciate sir .$
Commented by mrW1 last updated on 02/Jun/17

$g o o d w o r k i n g!$ $p l s f i n d t h e l a s t 3 d i g i t s .$
Commented by Tinkutara last updated on 02/Jun/17

$I do not know how to find last 3 digits .$ $Maybe in that case, \mod 1000 is$ $required . It will be very lengthy and$ $calculative (use of calculators) .$
Commented by RasheedSoomro last updated on 02/Jun/17

$e^{x} cellent!$
Commented by mrW1 last updated on 02/Jun/17

$v e r y g o o d!$ $p l e a s e t r y t o s o l v e$ $Q 13724$
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